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sergij07 [2.7K]
3 years ago
15

What is the range of the quadratic function?

Mathematics
1 answer:
tigry1 [53]3 years ago
4 0

For every polynomial function (such as quadratic functions for example), the domain is all real numbers. if the parabola is opening upwards, i.e. a > 0 , the range is y ≥ k ; if the parabola is opening downwards, i.e. a < 0 , the range is y ≤ k .

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Suppose the volume of milk in a glass is 230 cubic cm. If 1 gallon of milk equals 3.79 liters, how many glasses of milk are in 1
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C. 16 glasses (: 
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What is x, the angle at which the diagonal beam meets the 10-foot beam at the top of the frame? 16.7° 17.5° 72.5° 73.3°
umka21 [38]

The complete question is

"A theater production group is making frames to support wall-like props. Three-foot beams form right triangles with 10-foot beams to allow them to stand, as shown in the image.

What is x, the angle at which the diagonal beam meets the 10-foot beam at the top of the frame?

16.7°, 17.5°, 72.5°,73.3°"

The value of angle x would be 16.7 degrees.

<h3>What is the right triangle?</h3>

A right-angle triangle is a triangle that has a side opposite to the right angle the largest side and is referred to as the hypotenuse.

The angle of a right angle is always 90 degrees.

In order to find the value of x, a tan would be used.

This is because we have given the opposite and adjacent sides as 10 ft and 3 ft.

Tan Ф = opposite / adjacent

Tan Ф =  (3/10)

= 16.7 degrees

The value of angle x would be 16.7 degrees.

Learn more about a right angle;

brainly.com/question/7116550

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6 0
2 years ago
31<br>Write as a mixed number.<br>Give your answer in its simplest form.​
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3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
Consider the following system of two linear equations:
harkovskaia [24]

Hello,

Answer:

The last graph (2;3)

Step-by-step explanation:

2x + 3y = 12

2x – 3y = 0

2x + 3y = 12

2x = 3y

3y + 3y = 12 ⇔ 6y = 12 ⇔ y = 12/6 = 2

2x + 3 × 2 = 12 ⇔ 2x + 6 = 12 ⇔ 2x = 6 ⇔ x = 6/2 = 3

4 0
2 years ago
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